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sin+sincos+tan-tancos/1-cos^2sin+tan+sincos-sin/sin^2sincos+tan/sin^2sincos/sin^2+tan/sin^2cos/sin+1/sincoscot+sec.cosec

The bread, as well as the other products that are produced by making the use of the baker’s yeast, are definitely nutritious.

The yeast is meant to contain 21 calories per 7 grams. Since most of the single loaf bread preparations require only one envelope of the yeast, it does not lead to any major increase in the caloric content of the bread product.

Thus only a small quantity of yeast is needed to make fluffy bread without any calories.

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Since C = π - A - B, we want to maximize the function:

f(A,B) = cos(A) + cos(B) + cos(π-A-B)

f(A,B) = cos(A) + cos(B) - cos(A+B)

within the region R:

A+B<π0<A,B

Since it's an open region, we know the max cannot occur anywhere along the boundary of Rand must occur at some critical point(s) in the interior.

So we just have to find the point where the total derivative is 0. Of course, A and B aren't dependent on each other, so we can just set the two partials equal to zero to find the critical point(s):sin(A+B) - sin(A) = 0sin(A+B) - sin(B) = 0

So sin(A) = sin(B), thus A=B

But sin(A) = sin(2A) = 2sin(A)cos(A)

Thus cos(A) = 1/2

Note: we can divide by sin(A) because A≠0. This also excludes A=0 as a solution, since A=0 is in the boundary of R, which is not a part of our valid region.

A = π/3

So the max is at A=B=C=π/3.

Plugging that in, we know the maximum value of f is:

3cos(π/3) = 3/2

Thus for angles A,B,C of a triangle:

thank you

Let a be a given positive number.On dividing a by 4, let q be the quotient and r be the remainder.Then,by Euclid's algorithm,we have:a=4q+r where 0<=r<4a=4q+r where r=0,1,2,3a=4q+2=2(2q+1)It is clearly shown that 2q+1 is divisible by 2.Therefore,4q+2 is a positive integer.

thankssss

sin (90-A) = SEC a this is tegnometry formula

Sin (90-A)= cos Athanks

sin 90-A=CosA it is a trigonometry formula

cosA trigonometry basic question s

For compound interestA = P(1 + R/100)^n

Given, A = 12100, P = 10000, n = 2, R =?

Then,12100 = 10000(1 + R/100)^21.21 = (1 + R/100)^2(1 + R/100) = 1.1R/100 = 1.1 - 1 = .1R = .1*100 = 10

RATE OF INTEREST = 10%

10

that's alllllll

thank u

a) 11-7 = 4c) -10 +19 = 9e) -380-270 = -650

5.2 = 52/10 = 26/5

In p/q form its value = 26/5

Let a be any positive integer and b = 3a = 3q + r, where q ≥ 0 and 0 ≤ r < 3Therefore, every number can be represented as these three forms. There are three cases.

Case 1: When a = 3q,Where m is an integer such that m = (3q)3 = 27q39(3q3) = 9m

Case 2: When a = 3q + 1,a3= (3q +1)3a3= 27q3+ 27q2+ 9q + 1a3= 9(3q3+ 3q2+ q) + 1a3= 9m + 1Where m is an integer such that m = (3q3+ 3q2+ q)

Case 3: When a = 3q + 2,a3= (3q +2)3a3= 27q3+ 54q2+ 36q + 8a3= 9(3q3+ 6q2+ 4q) + 8a3= 9m + 8

Where m is an integer such that m = (3q3+ 6q2+ 4q)Therefore, the cube of any positive integer is of the form 9m, 9m + 1, or 9m + 8.

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6)26.87is greater than 35.63

62.50 - 35.63 = 26.87so by 26.87, 62.5 is greater than 35.63

6)26.87is greater than 35.63

62.50 - 35.63 = 26.87

26.87 is greater than 35.63

another type

Please write question in notebook and post

formula = principal amount *time *rate of interest/100

In the end of 3 yearsAmount will be 19965and interest will be19965-150004965

Final amount = 19965

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A + B + C = π ...... (1)L.H.S.

= ( cos A + cos B ) + cos C

= { 2 · cos[ ( A+B) / 2 ] · cos [ ( A-B) / 2 ] } + cos C

= { 2 · cos [ (π/2) - (C/2) ] · cos [ (A-B) / 2 ] } + cos C

= { 2 · sin( C/2 ) · cos [ (A-B) / 2 ] } + { 1 - 2 · sin² ( C/2 ) }

= 1 + 2 sin ( C/2 )· { cos [ (A -B) / 2 ] - sin ( C/2 ) }

= 1 + 2 sin ( C/2 )· { cos [ (A-B) / 2 ] - sin [ (π/2) - ( (A+B)/2 ) ] }

= 1 + 2 sin ( C/2 )· { cos [ (A-B) / 2 ] - cos [ (A+B)/ 2 ] }

= 1 + 2 sin ( C/2 )· 2 sin ( A/2 )· sin( B/2 ) ... ... ... (2)

= 1 + 4 sin(A/2) sin(B/2) sin(C/2)

= R.H.S.

Givenamount = 15000in first year ( 10 percentage of 15000)15000*10/100 because rate is 10 percentage1500new amount15000+150016500now again 10 percentage rate sonew amountwill be16500*10/100(10 percentage of 16500)165016500+165018150- new principal amountnow in third year18150*10/100( 10 percentage of the 18150)1815finally amount = 18150+181519965 rupeesand interest = 19965-150004965Answer

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The correct option is d) 2x + 3

Total weight of apples= 950 gm

Weight on other side = 500 gm

More weight he should put to weigh apples = 950 - 500= 450 gm

let ABC be a triangle in which the angle ABC is greater than the angle ACB.

TPT: AC > AB

proof:

let us try to prove this by contradiction. let us assume that AC is not longer than AB.

then two cases will arise:

case I : AC = AB

case II : AC < AB

if AC = AB. triangle ABC would has been isosceles triangle andangle ABC = angle BAC

this contradicts to the given condition.

in the II nd case: the side BC would has been longer than AC and consequently the. angle ABC < angle BAC.

but from the theorem " If in a triangle two sides are un equal , then angle opposite to longer side is greater than the angle opposite to shorter side".

again this contradicts the given condition.

thus the only remaining possibility is that the side AC is longer than the side BC.

thus AC > AB

Let the side opposite to the given angle be xx+90+60=180x+150=180 (Sum of the angles in a triangle x=180-150 is equal to 180.)x=30The side opposite to the given angle is 30 degrees.

A is the correct answer because supplementary means the sum of the angles is 90

Principal amount = Rs 75,000

Period = 2 years

Interest = 15% p.a.

Find the amount at the end of 1 year:

Interest = 15% x Rs 75,000 = 0.15 x 75000 = Rs 11,250

Total amount = 75000 + 11250 = Rs 86, 250

Find the amount at the end of 2 years:

Interest = 16% of Rs 86,250 = 0.15 x 86,250 = Rs 12,937

Total amount = 86, 250 + 12,937 = Rs 99,1 87

ok

thanks

1. 3 to the power 42. 8 to the power 93. 10 to the power 2

a)3^-4=3^4; b)8^-9=8^9, c)10^-3=10^3, d)67^-1=67^1; e)54^-5=54^4

By cosine law of triangles,cos(A) = (b² + c² - a²)/2bccos(B) = (a² + c² - b²)/2accos(C) = (a² + b² - c²)/2ab

iii) So, left side =(b² + c² - a²)/2abc + (a² + c² - b²)/2abc + (a² + b² - c²)/2abc

= (b² + c² - a² + a² + c² - b² + a² + b² - c²)/2abc = (a² + b² + c²)/2abc = Right side [Proved]

21 is the answer of your question